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First establish a comparison fact for a power series. If it converges at $w\ne 0$, its terms $a_n{w}^n$ are bounded, say $|a_n{w}^n|\le C$. For $|z|<|w|$,by the geometric series. Thus convergence at one point implies absolute convergence at every point of smaller modulus.

Let $S$ be the set of moduli of points at which the power series converges. It contains zero. Define the radius of convergence $R=\sup S\in [0,\infty ]$. If $|z|<R$, the definition of supremum gives a convergent point $w$ with $|w|>|z|$; the comparison just proved gives absolute convergence at $z$. This reasoning also applies when $R=\infty$. If $|z|>R$ and the power series converged there, its modulus would belong to $S$, contradicting the definition of $R$. ThereforeThe boundary $|z|=R$ is intentionally undecided by this argument. If $R=0$, only the exterior assertion is nonvacuous.

For the particular coefficients $a_n=1/n$ for $n\ge 1$, comparison with $\sum |z{|}^n$ gives absolute convergence whenever $|z|<1$. If $|z|=r>1$, then $r^n/n$ fails to tend to zero: its successive ratio is $rn/(n+1)$, eventually greater than a fixed number bigger than one. The necessary term condition for convergence of a series therefore fails. HenceFor instance the positive boundary point gives the divergent harmonic series; no assertion of universal boundary convergence is needed.

The two-variable arithmetic-geometric mean inequality follows from a square:Hence $2\sqrt{ab}\le a+b$, with equality exactly when $a=b$.

For the series, apply that same inequality with $a=a_n$ and $b=n^{-2}$ to obtainThe comparison series converges: its first summand has finite sum by hypothesis, and the inverse-square series converges, for example by the telescoping estimate $n^{-2}\le 1/[n(n-1)]=1/(n-1)-1/n$ for $n\ge 2$. Thus the comparison test for series provesEquivalently, its positive partial sums are increasing and bounded above by $\frac{1}{2}\sum a_n+\frac{1}{2}\sum n^{-2}$, so the bounded monotone sequence theorem proves convergence directly. This is the case $p=1$ of weighted square roots of a summable sequence.

To prove boundedness on $[a,b]$, suppose instead that $f$ is unbounded and choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence converging to $x_{\ast }\in [a,b]$. Continuity then gives $f(x_{n_j})\to f(x_{\ast })$, contradicting $|f(x_{n_j})|>n_j$. Hence $f$ is bounded.

Let $M={\sup }_{x\in [a,b]}f(x)$ and $m={\inf }_{x\in [a,b]}f(x)$. Choose $d_n$ with $M-1/n<f(d_n)\le M$. Another application of Bolzano-Weierstrass theorem yields a subsequence tending to $d\in [a,b]$, and continuity gives $f(d)=M$. Applying the same argument to $-f$ gives $c\in [a,b]$ with $f(c)=m$. ThusThis proves the extreme value theorem here, including both attainment assertions.

For $g$, its two limits supply $R>0$ such that $|g(x)|<1$ whenever $|x|>R$. On $[-R,R]$, continuity and the result just proved give a finite bound $K$. Consequently $|g(x)|\le \max \{K,1\}$ on the whole real line.

For the final value assertion, if $c=g(a)$, take $x=a$. Otherwise $0<c<g(a)$. The positive-infinity limit gives a point $b>a$ with $|g(b)|<c$, so $g(b)<c<g(a)$. The intermediate value theorem on $[a,b]$ now gives $x\in (a,b)$ withPositivity of $c$ is essential to this tail argument; a function can tend to zero without ever attaining zero.

For a uniform circular cross-section, averaging gives $⟨x^2⟩=⟨y^2⟩=R^2/4$; for a uniform axial coordinate in $[-h,h]$, $⟨z^2⟩=h^2/3$. Symmetry makes the coordinate axes principal axes, since all products of distinct coordinates integrate to zero. Thus the principal moments of inertia about the centre areThe inertia tensor is a scalar multiple of identity exactly when $h^2/3=R^2/4$. Thereforeis the height-to-radius ratio for sphere-like rotational inertia; the height-to-diameter ratio is $\sqrt{3}/2$. The geometric shape remains cylindrical, but its angular momentum is parallel to angular velocity for every axis, just as for a sphere.

For $\mathrm{Re}z>0$, absolute convergence permits the two-dimensional gamma integralPut $u=rs$, $v=r(1-s)$, with $r>0$, $0<s<1$ and Jacobian determinant $r$. Separation of the integrals gives the beta-gamma identityThe beta integral is finite in this half-plane. Thus a zero of $\Gamma (z)$ would imply $\Gamma (z/2)=0$, and iteration would imply $\Gamma (z/2^m)=0$ for every positive integer $m$. But integration by parts gives $w\Gamma (w)=\Gamma (1+w)$, and continuity near one makes $\Gamma (1+w)\to 1$ as $w\to 0$. It cannot vanish for sufficiently small $w$, a contradiction. Therefore the Gamma function has no zeros in the right half-plane.

For any $z$ outside the nonpositive integers, choose an integer $n$ with $\mathrm{Re}(z+n)>0$. The Gamma function recurrence givesAt the excluded integers gamma has poles, with nonzero residues by the same recurrence. Hence the meromorphic gamma function has no zeros anywhere; at its poles it has no finite value to which a nonzero-value assertion could apply.

The nonzero fixed point of the logistic map is $x_{\ast }=(\mu -1)/\mu$, with multiplier $F^{\prime }(x_{\ast })=2-\mu$. It is stable for $1<\mu <3$, and its multiplier crosses $-1$ at $\mu =3$. To identify the emerging orbit explicitly, factorThe remaining two roots, interchanged by $F$, areThey coincide with $x_{\ast }$ at three and are distinct for $\mu >3$. Their cycle multiplier isIt lies between $-1$ and $1$ for $3<\mu <1+\sqrt{6}$. This establishes the first period-doubling bifurcation and the stable bifurcating two-cycle.

At ${\mu }_2=1+\sqrt{6}$, ${\Lambda }_2=-1$ and $d{\Lambda }_2/d\mu =-2\sqrt{6}\ne 0$. The second flip is nondegenerate: for $H=F^2$ at either cycle point, the cubic flip coefficient is $c=(H^{\prime \prime \prime }/6)+(H^{\prime \prime }/2{)}^2=-\mathcal{S}H/6$ when $H^{\prime }=-1$. Here $\mathcal{S}$ denotes the Schwarzian derivative. Since $\mathcal{S}F=-6/(1-2x{)}^2<0$ and $\mathcal{S}(F^2)=(\mathcal{S}F\circ F)(F^{\prime }{)}^2+\mathcal{S}F$, this coefficient is positive. Locally, after following the fixed point of $H$, its second iterate has the form $H^2(y)-y=2a(\mu -{\mu }_2)y-2c{y}^3+\cdots$, with $a=2\sqrt{6}>0$. Nonzero roots therefore appear with $y^2\sim a(\mu -{\mu }_2)/c$ on the increasing-parameter side. These are a four-cycle of $F$, proving the second period-doubling bifurcation occurs at $\mu =1+\sqrt{6}$.